MCQ Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 4

Find two numbers whose sum is

$15$ and difference is

$3$ .

0%
$13$ and $2$
0%
$7$ and $8$
0%
$9$ and $6$
0%
$3$ and $12$

Explanation

Let the two numbers be

$x$ and

$y$ .

It is given that the sum of the numbers is

$15$ that is

$x+y=15..........(1)$

Also the difference of the numbers is

$3$ that is

$x-y=3..........(2)$ .

Add equations 1 and 2 to eliminate

$y$ , because the coefficients of

$y$ are opposite. So, we get

$(x+x)+(y-y)=15+3$

i.e.

$2x=18$

$\implies x=9$

Substituting this value of

$x$ in (1), we get

$9+y=15$

i.e.

$y=15-9=6$

Hence, the two numbers are

$9$ and

$6$ .

Solve the following pair of simultaneous equations:

$\displaystyle \frac{a}{4}\, -\, \frac{b}{3}\, =\, 0\,;\, \frac{3a\, +\, 8}{5}\, =\, \frac{2b\, -\, 1}{2}$

0%
$a = -4.5, b = 12$
0%
$a = 4, b = -5$
0%
$a = 14, b = 10.5$
0%
$a = 12, b = 11.5$

Explanation

$\dfrac { a }{ 4 } -\dfrac { b }{ 3 } =0\\ \Rightarrow a =\dfrac { 4b }{ 3 }$

Substituting for '

$a$ ' in the other equation.

$\dfrac { 3a+8 }{ 5 } =\dfrac { 2b-1 }{ 2 } \\ 6a+16 =10b - 5\\ 6*\dfrac { 4b }{ 3 } + 16 = 10b - 5\\ 8b +16 = 10b - 5\\ 21 = 2b\\ b = 10.5$

So,

$a = \dfrac { 4\times10.5 }{ 3 } = 14$

$\therefore a = 14, b = 10.5$

Option

$C$

Solve the following pair of equations :

$x-y= 0.9$ and

$\displaystyle \frac{11}{2\left ( x+y \right )}= 1$

0%
$x= 1.5;y= 4$
0%
$x= 3;y= 2.5$
0%
$x= 5.2;y= 0.3$
0%
$x= 3.2;y= 2.3$

Explanation

$x=y+0.9$

Substituting in the 2nd equation:

$11=2(y+0.9+y)$

$\dfrac{11}{2}=2y+0.9\\5.5-0.9=2y\\4.6=2y\\ \therefore y=2.3$

and

$x=2.3+0.9=3.2$

Therefore,

$x=3.2$ and

$y=2.3$

Find a fraction which reduces to

$\displaystyle \frac{2}{3}$ if the numerator and the denominator are each increased by

$1$ , and reduces to

$\displaystyle \frac{3}{5}$ if the numerator and the denominator are each decreased by

$2$ .

0%
$\displaystyle \frac{13}{12}$
0%
$\displaystyle \frac{11}{17}$
0%
$\displaystyle \frac{9}{4}$
0%
$\displaystyle \frac{13}{3}$

Explanation

Let the fraction be

$\dfrac {x}{y}.$

Given, if the numerator and the denominator are each increased by

$1$ , then fraction

$= \dfrac {x + 1}{y + 1} = \dfrac {2}{3}$

$\Longrightarrow 3x + 3 = 2y + 2$

$\Longrightarrow 3x - 2y = -1$ .......................

$(1)$

Also, if the numerator and the denominator are each decreased by

$2$ , then fraction

$= \dfrac {x - 2}{y - 2} = \dfrac {3}{5}$

$\Longrightarrow 5x - 10 = 3y - 6$

$\Longrightarrow 5x - 3y = 4$ .......................

$(2)$

Multiplying equation

$(1)$ with

$5$ we get,

$15x - 10y = -5$ ...

$(3)$

Multiplying equation $(2)$ with $3$ we get, $15x - 9y = 12$ .... $(4)$

Subtracting equations $(3)$ from $(4)$ , we get,

$-9y-(-10y) = 12-(-5)$

$y=17$

Substituting $y = 17$ in the equation $(2)$ , we get,

$5x - 3(17) = 4$

$5x = 55$

$\Longrightarrow x = 11$

Hence, the fraction is $\dfrac {11}{17}.$

Find two numbers, which differ by

$7$ , such that twice the greater added to five times the smaller gives

$42$ .

0%
$19$ and $12$
0%
$12$ and $5$
0%
$11$ and $4$
0%
$16$ and $9$

Explanation

Let the two numbers be

$x$ and

$y$ .

It is given that two numbers differ by

$7$

$\Rightarrow x-y=7$

$...(1)$

Also, twice the greater number added to five times the smaller gives

$42$

$\Rightarrow 2x+5y=42$

$...(2)$

Multiply equation

$(1)$ by

$2$

$2x-2y=14$

$...(3)$

Let us subtract equation

$(3)$ from equation

$(2)$ to eliminate

$x$ , because the coefficients of

$x$ are the same.

$(2x+5y)-(2x-2y)=42-14$

$\Rightarrow (2x-2x)+(5y+2y)=28$

$\Rightarrow 7y=28$

$\Rightarrow y=4$

Substituting this value of

$y$ in

$(1),$

we get

$x-4=7$

i.e.

$x=11$

Hence, the two numbers are

$11$ and

$4$

$A$ is

$25$ years older than

$B$ . In

$15$ years,

$A$ will be twice of

$B$ . Find the present ages of

$A$ and

$B$ .

0%
Present age of $A$ is $40$ years and

Present age of $B$ is $15$ years
0%
Present age of $A$ is $37$ years and

Present age of B is $12$ years
0%
Present age of $A$ is $35$ years and

Present age of $B$ is $10$ years
0%
Present age of $A$ is $45$ years and

Present age of $B$ is $20$ years

Explanation

Let the present age of

$A$ be

$x$ and the present age of

$B$ be

$y$ .

It is given that

$A$ is

$25$ years older than

$B$ that is:

$x=y+25$

$x-y=25..........(1)$

Also, In

$15$ years,

$A$ will be twice of

$B$ that is:

$x+15=2(y+15)$

$x+15=2y+30$

$x-2y=30-15$

$x-2y=15..........(2)$

Now subtract equation 1 from equation 2 to eliminate

$x$ , because the coefficients of

$x$ are same. So, we get

$(x-x)+(-2y+y)=15-25$

i.e.

$-y=-10$

i.e.

$y=10$

Substituting this value of

$y$ in (1), we get

$x-10=25$

i.e.

$x=25+10=35$

Hence, the present age of

$A$ is

$35$ years and the present age of

$B$ is

$10$ years.

A man is 24 years older than his son. 12 years ago, he was five times as old as his son. Find the present ages of both.

0%
Present age of father is $44$ years and Present age of son is $20$ years
0%
Present age of father is $42$ years and Present age of son is $18$ years
0%
Present age of father is $60$ years and Present age of son is $36$ years
0%
Present age of father is $48$ years and Present age of son is $24$ years

Explanation

Let the present age of the father be

$x$ years and the present age of the son be

$y$ years.

It is given that the man is

$24$ years older than his son that is:

$x=y+24$

$\Rightarrow x-y=24\quad ..........(1)$

Also,

$12$ years ago, he was five times as old as his son that is:

$(x-12)=5(y-12)$

$\Rightarrow x-12=5y-60$

$\Rightarrow x-5y=-60+12$

$\Rightarrow x-5y=-48\quad .......(2)$

On subtracting

$(1)$ from

$(2)$ , we get:

$(x-x)+(-5y+y)=-24-48$

$\Rightarrow -4y=-72$

$\Rightarrow y=18$

Substituting this value of

$y$ in

$(1)$ , we get:

$x-18=24$

$\Rightarrow x=24+18=42$

Hence, the present age of the father is

$42$ years and the present age of the son is

$18$ years.

Find the fraction such that it becomes

$\displaystyle \frac{1}{2}$ if 1 is added to the numerator, and

$\displaystyle \frac{1}{3}$ if 1 is added to the denominator.

0%
$\dfrac{3}{8}$
0%
$\dfrac{1}{6}$
0%
$\dfrac{8}{7}$
0%
$\dfrac{7}{3}$

Explanation

Let the fraction be

$\dfrac { x }{ y }$

$\therefore \dfrac { x+1 }{ y } = \dfrac { 1 }{ 2 }$ ...(1)

$\dfrac { x }{ y+1 } = \dfrac { 1 }{ 3 }$ ...(2)

So from

$1$ we get,

$2x+2 = y$ ...(3)

From

$2$ we get,

$3x = y+1$ ...(4)

Substitute

$eq(3)$ in

$eq(4)$ , we get

$3x = 2x+2+1$

$x = 3\\ y = 2x+2 = 8$

The fraction is

$\dfrac { 3 }{ 8 }$

Solve the following pair of linear (simultaneous) equations by the method of elimination:

$2x+3y= 8$

$2x= 2+3y$

0%
$x= 3$ and $y=6$
0%
$x= 5.5$ and $y=-4$
0%
$x= 3$ and $y=2.5$
0%
$x= 2.5$ and $y=1$

Explanation

$2x + 3y = 8$ ...

$(1)$

$2x - 3y = 2$ ...

$(2)$

Adding

$(1)$ and

$(2)$ ,

$4x = 10$

$x = \dfrac{1}{4}$ or

$x = \dfrac{5}{2}$

Substituting

$x = \dfrac{5}{2}$ in

$(1)$ , we get

$5 + 3y = 8$

$3y = 3$

$y = 1$

So,

$x = \dfrac{5}{2}, y = 1$

Solve the following pair of linear (simultaneous) equations by the method of elimination:

$x+8y= 19$ ,

$2x+11y= 28$

0%
$x= 3$ and $y= 2$
0%
$x= 12$ and $y=5$
0%
$x= 1$ and $y=6$
0%
$x= 9$ and $y=-6$

Explanation

Multiply the equation

$x+8y=19$ by

$2$ to make the coefficients of

$x$ equal. Then we get the equations:

$2x+16y=38.........(1)$

$2x+11y=28.........(2)$

Subtract Equation (2) from Equation (1) to eliminate

$x$ , because the coefficients of

$x$ are the same. So, we get

$(2x-2x)+(16y-11y)=38-28$

i.e.

$5y =10$

i.e.

$y = 2$

Substituting this value of

$y$ in the equation

$x+8y=19$ , we get

$x+16=19$

i.e.

$x=19-16=3$

Hence, the solution of the equations is

$x = 3, y = 2$ .

Solve the following pair of linear (simultaneous) equations by the method of elimination:

$2x+7y= 39$

$3x+5y= 31$

0%
$x= 1,y= 6$
0%
$x= 7,y= 4$
0%
$x= 2,y= 5$
0%
$x= 3,y= 1$

Explanation

$2x + 7y = 39$ ...

$(1)$

$3x + 5y = 31$ ...

$(2)$

Multiplying

$(1)$ by

$3$ and

$(2)$ by

$2$ ,

$6x + 21y = 117$ ...

$(3)$

$6x + 10y = 62$ ...

$(4)$

Subtracting

$(3)$ and

$(4)$ ,

$11y = 55$

$y = 5$

Substituting

$y = 5$ in

$(1)$ , we get

$2x + 35 = 39$

$2x = 4$

$x = 2$

So,

$x = 2, y = 5$

Option

$C$

Solve the following pair of linear (simultaneous) equations by the method of elimination:

$2x-3y= 7$

$5x+y= 9$

0%
$x= 4$ and $y=1$
0%
$x= 1$ and $y=-2$
0%
$x= 2$ and $y=-1$
0%
$x= 5$ and $y=-4$

Explanation

$2x - 3y = 7$ ...

$(1)$

$5x + y = 9$ ...

$(2)$

Multiplying

$(2)$ by

$3$ and adding

$(1)$ and

$(3)$ ,

$2x - 3y = 7$ ...

$(1)$

$15x + 3y = 27$ ...

$(3)$

$17x = 34$

$x = 2$

Substituting

$x = 2$ in

$(1)$ , we get

$4 - 3y = 7$

$-3y = 3$

$y = - 1$

So,

$x = 2, y = -1$

Solve the following pair of linear (simultaneous) equations by the method of elimination:

$8x+5y= 9$

$3x+2y= 4$

0%
$x= 1$ and $y=-2$
0%
$x= 1$ and $y=-5$
0%
$x= 3$ and $y=6$
0%
$x= -2$ and $y=5$

Explanation

$8x + 5y = 9$ ...

$(1)$

$3x + 2y = 4$ ...

$(2)$

Multiplying

$(1)$ by

$3$ and

$(2)$ by

$8$ ,

$24x + 15y = 27$ ...

$(3)$

$24x + 16y = 32$ ...

$(4)$

Subtracting

$(3)$ and

$(4)$ ,

$-y = -5$ or

$y=5$

Substituting

$y = 5$ in

$(1)$ , we get

$8x + 25 = 9$

$8x = -16$

$x = -2$

$x = -2, y = 5$

Option

$D$

Solve the following pair of linear (simultaneous) equations by the method of elimination :

$x+y= 7$

$5x+12y= 7$

0%
$x= 11$ and $y=-4$
0%
$x= 1$ and $y=7$
0%
$x= 13$ and $y=6$
0%
$x= 1$ and $y=-3$

Explanation

$x+y=7$ .... (1)

$5x+12y=7$ ....(2)

Multiply eq. (1) by 5, we get

$5x+5y=35$ ....(3)

Subtracting eq(2) from eq.(3)

$5x+5y=35$

$5x+12y=7$

$-$

---------------------

$-7y=28$

$y=-4$

Substituting value of

$y$ in eq,(1), we get,

$x-4=7$

$x=11$

Solve the following pair of linear (simultaneous) equations by the method of elimination:

$6x= 7y+7$

$7y-x= 8$

0%
$\displaystyle x= 3,\displaystyle y= \displaystyle \frac{11}{7}$
0%
$\displaystyle x= 2,\displaystyle y= \displaystyle \frac{1}{2}$
0%
$\displaystyle x= 6,\displaystyle y= \displaystyle \frac{13}{5}$
0%
$\displaystyle x= 4,\displaystyle y= \displaystyle \frac{-1}{2}$

Explanation

$6x - 7y = 7$ ...

$(1)$

$-x + 7y = 8$ ...

$(2)$

Adding

$(1)$ and

$(2)$ ,

$5x = 15$

$x = 3$

Substituting

$x = 3$ in

$(1)$ , we get

$18 - 7y = 7$

$7y = 11$

$y = \dfrac{11}{7}$

So,

$x =3, y = \dfrac{11}{7}$

Option

$A$

Solve the following pair of linear (simultaneous) equations by the method of elimination :

$1.5x+0.1y= 6.2$

$3x-0.4y= 11.2$

0%
$x= 4,y= 2$
0%
$x= 2,y= 5$
0%
$x= 3,y= 7$
0%
$x= 1,y= 6$

Solve the following pair of linear (simultaneous) equations by the method of elimination:

$0.2x+0.1y= 25$

$2\left ( x-2 \right )-1.6y= 116$

0%
$\displaystyle x= 50,\displaystyle y= -75$
0%
$\displaystyle x=-29,\displaystyle y= 66$
0%
$\displaystyle x= 100,\displaystyle y= 50$
0%
$\displaystyle x= 60,\displaystyle y= 30$

Explanation

$0.2x + 0.1y = 25$ ...

$(1)$

$2x - 1.6y = 120$ ...

$(2)$

Multiplying

$(1)$ by

$10$ and subtracting

$(3)$ and

$(2)$ ,

$2x + y = 250$ ...

$(3)$

$2x - 1.6y = 120$ ...

$(2)$

$2.6y = 130$

$y = 50$

Substituting

$y = 50$ in

$(1)$ , we get

$0.2x + 5 = 25$

$0.2x = 20$

$x = 100$

So,

$x = 100, y = 50$

Option

$C$

Solve the following pair of equations:

$7x+6y= 71$

$5x-8y= -23$

0%
$x= -5;y= 4$
0%
$x= 3,y= -2$
0%
$x= 5,y= 6$
0%
$x= 1,y= 3$

Explanation

$\\ 7x+6y=71;\\ y=\dfrac{71-7x}{6};\\$

Substituting:

$5x-8\left(\dfrac{71-7x}{6}\right)=-23$

$\Rightarrow 30x-568+56x=-138$

$\Rightarrow 86x=430$

$\Rightarrow x=5$

Resubstituting the value of

$x$ , we get

$y=\dfrac{71-35}{6}=6$

Therefore

$x=5, y=6$

Solve the following pair of equations using substitution method:

$\displaystyle \frac{5y}{2}-\displaystyle \frac{x}{3}= 8$

$\displaystyle \frac{y}{2}+\displaystyle \frac{5x}{3}= 12$

0%
$x= 2;y= 7$
0%
$x= 6;y= 4$
0%
$x= 7;y= 1$
0%
$x= 8;y= 3$

Explanation

$\dfrac { 5y }{ 2 } -\dfrac { x }{ 3 } =8\quad ...(1)\\ \dfrac { y }{ 2 } +\dfrac { 5x }{ 3 } =12\quad ...(2)$

Let

$\dfrac{y}{2} = a$ and

$\dfrac{x}{3} = b$

So, the original equations reduce to

$5a - b = 8$ ...

$(1)$

$a + 5b = 12$ ...

$(2)$

From

$(1), b = 5a - 8$

Substituting

$b = 5a - 8$ in

$(2)$ , we get

$a + 5(5a - 8) = 12$

$a + 25a - 40 = 12$

$26a = 52$

$a = 2$

So,

$b = 5a - 8$

$b = 10 - 8 = 2$

But

$\dfrac{y}{2} = a$ or

$y = 2a = 4$

$\dfrac{x}{3} = b$ or

$x = 3b = 6$

$x = 6, y = 4$

Option

$B$ .

Solve the following equations by substitution method.

$2x+y=-2;\, \, 3x-y=7$

0%
$x = 1, y= -4$
0%
$x = 1, y= -1$
0%
$x = 1, y= -3$
0%
$x = 1, y= -5$

Explanation

$2x+y =-2; \, \, 3x-y =7$

$y = -2x -2$

Put it in second equation.

$3x -(-2x-2) =7$

$5x +2 = 7$

$x =1$

$y = -2x-2 = -2 (1)-2 = -4$

$x =1$ and

$y = -4$

Solve the following pair of equations:

$3-\left ( x-5 \right )= y+2$ ,

$2\left ( x+y \right )= 4-3y$

0%
$x= \displaystyle \frac{7}{2},y= -\displaystyle \frac{9}{5}$
0%
$x= \displaystyle \frac{1}{6},y= -\displaystyle \frac{4}{3}$
0%
$x= \displaystyle \frac{26}{3},y= -\displaystyle \frac{8}{3}$
0%
$x= \displaystyle \frac{6}{5},y= -\displaystyle \frac{7}{3}$

Explanation

The equation

$3-(x-5)=y+2$ can be solved as:

$3-x+5=y+2\\ \Rightarrow 8-x=y+2\\ \Rightarrow x+y=6\quad .........(1)$

The equation

$2(x+y)=4-3y$ can be solved as:

$2x+2y=4-3y\\ \Rightarrow 2x+2y+3y=4\\ \Rightarrow 2x+5y=4\quad .........(2)$

Multiply the equation 1 by

$2$ . Then we get:

$2x+2y=12.........(3)$

Subtract equation 3 from equation 4 to eliminate

$x$ , because the coefficients of

$x$ are the same. So, we get

$(2x-2x)+(5y-2y)=4-12$

i.e.

$3y=-8$

i.e.

$y=-\dfrac { 8 }{ 3 }$

Substituting this value of

$y$ in the equation 1, we get

$x+\left( -\dfrac { 8 }{ 3 } \right) =6\\ \Rightarrow 3x-8=18\\ \Rightarrow 3x=18+8=26\\ \Rightarrow x=\dfrac { 26 }{ 3 }$

Hence, the solution of the equations is

$x=\dfrac { 26 }{ 3 } ,y=-\dfrac { 8 }{ 3 }$ .

Solve the following pair of equations:

$\displaystyle \frac{1}{5}\left ( x-2 \right )=\displaystyle \frac{1}{4}\left ( 1-y \right )$ ,

$26x+3y+4= 0$

0%
$x= \displaystyle -\frac{1}{2};y= 3$
0%
$x= \displaystyle -\frac{6}{5};y= 9$
0%
$x= \displaystyle -\frac{8}{5};y= 8$
0%
$x= \displaystyle -\frac{2}{3};y= 7$

Explanation

The equation

$\dfrac { 1 }{ 5 }(x-2) =\dfrac { 1 }{ 4 }(1-y)$ can be rewritten as:

$\dfrac { 1 }{ 5 } (x-2)=\dfrac { 1 }{ 4 } (1-y)$

$\Rightarrow 4(x-2)=5(1-y)\\ \Rightarrow 4x-8=5-5y\\ \Rightarrow 4x+5y=5+8\\ \Rightarrow 4x+5y=13\quad .........(1)$

The equation

$26x+3y+4=0$ can be rewritten as:

$26x+3y=-4..........(2)$

Multiply the equation 1 by

$3$ and equation 2 by

$5$ to make the coefficients of

$y$ equal. Then we get the equations:

$12x+15y=39.........(3)$

$130x+15y=-20.........(4)$

Subtract Equation 4 from equation 3 to eliminate

$y$ , because the coefficients of

$y$ are the same. So, we get

$(12x-130x)+(15y-15y)=39+20$

i.e.

$-118x=59$

i.e.

$x=-\dfrac { 1 }{ 2 }$

Substituting this value of

$x$ in the equation 1, we get

$-2+5y=13$

i.e.

$5y=13+2=15$

i.e.

$y=3$

Hence, the solution of the equations is

$x=-\dfrac { 1 }{ 2 },y=3$ .

The sides of an equilateral triangle are given by

$x+3y$ ;

$3x+2y-2$ and

$4x+\displaystyle \frac{1}{2}y+1$ respectively. Find the lengths of the sides of the triangle.

0%
$15$ units each
0%
$17$ units each
0%
$12$ units each
0%
$11$ units each

Explanation

Since it is an equilateral triangle, the lengths of the sides are equal

So,

$x+3y = 3x+2y-2 = 4x+\dfrac { 1 }{ 2 } y+1$

$\therefore x + 3y = 3x+2y -2\\ \therefore y = 2x - 2$

$2x - y = 2$ ...(1)

Also,

$4x+\dfrac { 1 }{ 2 } y+1 = x+3y$

$\therefore 8x + y + 2 = 2x +6y\\ 6x + 2 = 5y$

$6x - 5y = -2$ ...(2)

Multiplying eq(1) by 5 we get,

$10x - 5y = 10$ ...(3)

Subtracting eq(2) from eq(3) we get,

$4x = 12 \Rightarrow x = 3$

$\therefore y = 2x -2 = 4$

The length of one side of equilateral triangle =

$x +3y$ =

$3 + 3(4) = 15$ units

Solve the following equations by substitution method.

$2x-3y-3=7; \, \, 4x-5y-5=10$

0%
$x = \displaystyle \frac{-7}{2}, y= -5$
0%
$x = \displaystyle \frac{-3}{2}, y= -5$
0%
$x = \displaystyle \frac{-1}{2}, y= -5$
0%
$x = \displaystyle \frac{-5}{2}, y= -5$

Explanation

Rewriting the given equations, we get

$2x-3y=10$ ...(i)

$4x-5y=15$ ...(ii)
We pick either of the equations and write one variable in terms of the other. Consider the Equation (i) and write it as

$x=\displaystyle \frac {10+3y}{2}$ ...(iii)
Substitute the value of

$x=\displaystyle \frac {10+3y}{2}$ in Equation (ii). we get,

$4\left(\displaystyle \dfrac {10+3y}{2}\right)-5y=15$

$\Rightarrow 20+6y-5y=15$

$\Rightarrow y=-5$
Substituting this value of

$y=-5$ in Equation (iii). we get,

$x=\displaystyle \frac {10+3(-5)}{2}= \frac {-5}{2}$

$\therefore x=\dfrac {-5}{2} , y=-5$

Solve the following pair of equations:

$2x-3y-3= 0$ ,

$\displaystyle \frac{2x}{3}+4y+\displaystyle \frac{1}{2}= 0$

0%
$x= \displaystyle \frac{13}{20},y= -\displaystyle \frac{7}{10}$
0%
$x= \displaystyle \frac{17}{20},y= -\displaystyle \frac{1}{20}$
0%
$x= \displaystyle \frac{1}{10},y= -\displaystyle \frac{7}{10}$
0%
$x= \displaystyle \frac{21}{20},y= -\displaystyle \frac{3}{10}$

Explanation

$2x-3y-3= 0$ ...... (1)

$\displaystyle \frac{2x}{3}+4y=-\displaystyle \frac{1}{2}$ ...... (2)

$2x-3y=3;$

$x=\dfrac{3+3y}{2}$ .....(3)

Substituting in the 2nd equation

$\dfrac{-1}{2}=\dfrac{2}{3} \times \dfrac{3+3y}{2} +4y$

$5y=-\dfrac{3}{2} \implies y=\dfrac{-3}{10}$

Put the value of

$y$ in (3)

$x=\dfrac{3+3(\dfrac{-3}{10})}{2} \implies x=\dfrac{21}{20}$

$\therefore x=\dfrac{21}{20} \quad and \quad y=\dfrac{-3}{10}$

Pooja and Ritu can do a piece of work in

$\displaystyle 17\frac{1}{7}$ days. If one day work of Pooja is three fourth of one day work of Ritu then find in how many days each alone will do the same work.

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Pooja in 40 days and Ritu in 30 days
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Pooja in 60 days and Ritu in 40 days
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Pooja in 30 days and Ritu in 10 days
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Pooja in 45 days and Ritu in 20 days

Explanation

Let the work done by pooja in one day be

$1/x$ and work done by Ritu in one day be

$1/y.$

Therefore, according to the question total work done by both in one day,

$\\ 1/x+1/y=\dfrac{1}{17\dfrac{1}{7}} = 7/120$ .....(1)

Also, one day work of Pooja is three fourth of one day work of Ritu

$\\ 1/x=3/4y$

Substituting above equation in equation (1) , we get

$y=40$

Substituting

$y=40$ in equation (1) we get

$x=30$
Hence pooja will complete the work in

$40$ days whereas ritu in

$30$ days

Solve the following equations by substitution method.

$x-2y+2=0; \, \, x+2y=10$

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$x = 4, y= 3$
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$x = 1, y= 3$
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$x = 9, y= 3$
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$x = 5, y= 3$

Explanation

$x-2y+2=0$ ....(i)

$x+2y=10$ ....(ii)
We pick either of the equations and write one variable in terms of the other.
Let us consider the Equation (i) and write it as

$x=2y-2$ .........(iii)
Substitute the value of

$x=2y-2$ in Equation (ii). We get

$2y-2+2y=10$

$\Rightarrow 4y=12$

$\Rightarrow y=3$
Substituting this value of

$y=3$ in Equation (iii), we get

$x=2(3)-2=4$

$\therefore x=4 , y=3$

Solve the following simultaneous equations by the method of equating coefficients.

$3x-4y=7;\, \, 5x+2y=3$ .

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$x = 1, y= -1$
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$x = 2, y= 1$
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$x = 5, y= 2$
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$x = 4, y= -3$

Explanation

$3x-4y=7$ ...(i)

$5x+2y=3$ ...(ii)
On multiplying (i) by 5 and (ii) by 3, we get

$15x-20y=35$

$\underline {\underset {-}15x\underset {-}{+}6y=\underset{-}9}$

$-26y=26$

$\therefore y=-1$
On putting

$y=-1$ in (i), we get

$3x-4(-1)=7$

$x=1$

$\therefore x=1,y=-1$

Solve the following simultaneous equations by the method of equating coefficients.

$\displaystyle \frac{x}{3} + \frac{y}{4} = 4; \, \, \frac{x}{2} - \frac{y}{4}=1$

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$x = 2, y= 8$
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$x = 6, y= 8$
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$x = 7, y= 8$
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$x = 1, y= 8$

Explanation

Given,

$\displaystyle \frac{x}{3} + \frac{y}{4} = 4$ ....(i)

$\displaystyle \frac{x}{2} - \frac{y}{4}=1$ .....(ii)

Multiplying equation (i) by

$12$ and equation (ii) by

$4$ . we get,

$4x+3y=48$ .......(iii)

$2x-y=4$ .....(iv)

Multiplying equation (iv) by

$3$ . we get,

$6x-3y=12$ .......(v)

Now, Adding (iii) and (v). we get,

$10x=60 \Rightarrow x=6$

Substitute

$x=6$ in equation (iii). we get,

$4(6)+3y=48$

$\Rightarrow 3y=24 \Rightarrow y=8$

$\therefore x=6 , y=8$

Solve the following equations by substitution method:

$5x-2y=13; \, \, 4x+3y=15$

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$x=1,y=-3$
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$x=3,y=1$
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$x=-1,y=3$
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$x=-1,y=-3$

Explanation

We pick either of the equations and write one variable in terms of the other.

Let us consider the equation

$5x-2y=13$ and write it as

$x=\frac { 13+2y }{ 5 }$

Substitute the value of

$x$ in equation

$4x+3y=15$ . We get

$4\left( \frac { 13+2y }{ 5 } \right) +3y=15\\ \Rightarrow 52+8y+15y=75\\ \Rightarrow 23y=23\\ \Rightarrow y=1$

Therefore,

$y = 1$

Substituting this value of

$y$ in the equation

$5x-2y=13$ , we get

$5x-2=13$

$5x=13+2=15$

$x=3$

Hence, the solution is

$x = 3, y = 1$ .

CBSE Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 4 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 4 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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